# How do you solve tan(arccos((sqrt2)/2))?

Jul 18, 2016

Ans.$= 1$.

#### Explanation:

Recall that $\arccos x = \theta , | x | \le 1 \iff \cos \theta = x , \theta \in \left[o , \pi\right]$

Hence, $\arccos \left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$, and, so,

$\tan \left\{\arccos \left(\frac{\sqrt{2}}{2}\right)\right\} = \tan \left(\frac{\pi}{4}\right) = 1$.

Jul 18, 2016

$\tan \left[\arccos \left(\frac{\sqrt{2}}{2}\right)\right] = 1$

#### Explanation:

$\arccos \left(\frac{\sqrt{2}}{2}\right)$ returns the angle $\theta$ where we have:

$\cos \left(\theta\right) = \left(\text{adjacent")/("hypotenuse}\right) = \frac{\sqrt{2}}{2}$
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But we need the tangent and this means we need to know the length of the opposite (h).

$\implies h = \sqrt{{2}^{2} - 2} = \sqrt{2}$

So $\tan \left[\arccos \left(\frac{\sqrt{2}}{2}\right)\right] \to \tan \left(\theta\right) = \frac{h}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1$